Step 1: Convert the critical t value into a critical correlation coefficient.
Instead of computing $t$ from $R$, rearrange the test statistic formula to find the smallest $|R|$ that would be judged significant at the given critical value $t_{0.025,60}=2.0$: \[ t = \frac{R\sqrt{n-2}}{\sqrt{1-R^2}} \implies R_{crit} = \frac{t}{\sqrt{t^2 + (n-2)}} \]
Step 2: Substitute the numbers.
With $t = 2.0$ and $n-2=60$: $t^2 = 4$, so \[ R_{crit} = \frac{2.0}{\sqrt{4+60}} = \frac{2.0}{\sqrt{64}} = \frac{2.0}{8} = 0.25 \] Any sample correlation with $|R| \geq 0.25$ in magnitude will give a test statistic at least as large as the critical value of 2.0.
Step 3: Compare the observed $R$ to this critical value.
The observed correlation is $R = 0.27$, and $0.27 > 0.25 = R_{crit}$. Since the observed correlation exceeds the critical correlation, the test statistic must exceed $t_{0.025,60}=2.0$, so the null hypothesis of zero correlation is rejected at $\alpha=0.05$.
Step 4: State the conclusion and rule out the distractors.
Because $H_0$ is rejected, $X$ and $Y$ show a statistically significant linear correlation, confirming option (A). Statistically independent (B) would require $R$ to be essentially zero and non significant, the opposite of what is found. Not significantly correlated (C) is the conclusion reached only if $R$ had been below 0.25. Physically related (D) is outside the scope of a correlation significance test, which only addresses statistical association.
\[ \boxed{\text{Option (A) is correct: significant linear correlation}} \]